Theorem elsucg 2290

Description: Membership in a successor. Exercise 5 of [TakeutiZaring] p. 17.

Assertion

Ref	Expression
elsucg	⊢ (A ∈ C → (A ∈ suc B ↔ (A ∈ B ∨ A = B)))

Proof of Theorem elsucg

Step	Hyp	Ref	Expression
1		elsncg 1825	. . 3 ⊢ (A ∈ C → (A ∈ {B} ↔ A = B))
2	1	orbi2d 466	. 2 ⊢ (A ∈ C → ((A ∈ B ∨ A ∈ {B}) ↔ (A ∈ B ∨ A = B)))
3		df-suc 2205	. . . 4 ⊢ suc B = (B ∪ {B})
4	3	eleq2i 1153	. . 3 ⊢ (A ∈ suc B ↔ A ∈ (B ∪ {B}))
5		elun 1601	. . 3 ⊢ (A ∈ (B ∪ {B}) ↔ (A ∈ B ∨ A ∈ {B}))
6	4, 5	bitr 151	. 2 ⊢ (A ∈ suc B ↔ (A ∈ B ∨ A ∈ {B}))
7	2, 6	syl5bb 410	1 ⊢ (A ∈ C → (A ∈ suc B ↔ (A ∈ B ∨ A = B)))

Syntax hints:   → wi 2   ↔ wb 127   ∨ wo 195   = wceq 1091   ∈ wcel 1092   ∪ cun 1485  {csn 1808  suc csuc 2201

This theorem is referenced by:  elsuc 2292  elelsuc 2295  ordsssuc 2310  ordsucelsuc 2324  suc11reg 3456  nlt1pi 3827

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-un 1490  df-sn 1811  df-pr 1812  df-suc 2205

metamath.org